What happens? At each stage we take the rightmost term and factor it into: ; continuing this way gives the above factorization of .

Now we can write (1) in the following form: (2).

Suppose that . Then by (2), , where is a positive integer. It follows that any common divisor of and must divide , and therefore is equal to or . Fermat Numbers are odd, so the only common divisor they have is .

Incidentally, the result of this problem is one of the proofs that there are infinitely many prime numbers....